|
%I #43 Mar 12 2019 18:30:58
%S 1,2,1,4,2,1,3,8,2,1,5,3,6,2,1,16,8,3,9,2,1,4,11,7,3,5,2,1,14,4,15,32,
%T 3,7,2,1,18,8,4,6,20,3,21,2,1,10,23,15,4,8,6,3,26,2,1,5,7,13,29,4,30,
%U 14,3,64,2,1,33,5,9,7,35,4,36,17,3,6,2,1,39,14,5,19,41,7,4,20,12,3,44,2,1,8
%N Smallest positive number starting an interval of consecutive integers with element sum n.
%C Right border of A299765. - _Omar E. Pol_, Jul 24 2018
%C In other words: a(n) is smallest part of the partitions of n into consecutive parts. - _Omar E. Pol_, Mar 12 2019
%H Alois P. Heinz, <a href="/A118235/b118235.txt">Table of n, a(n) for n = 1..20000</a> (first 1000 terms from Paul D. Hanna)
%F A109814(n) * (A109814(n) + 2*a(n) - 1) / 2 = n.
%F a(m) = n iff m = 2^k: a(A000079(n)) = A000079(n);
%F a(m) = 1 iff m = k*(k+1)/2: a(A000217(n)) = 1.
%F a(A002817(n-1)+1) = n; i.e., a(m) = n if m = k*(k-1)/2 + 1 and k = n*(n-1)/2 + 1. - _Paul D. Hanna_, Oct 28 2011
%F a(m) = 2 iff m = k*(k+3)/2: a(A000096(n)) = 2. - _Bernard Schott_, Mar 12 2019
%e a(3)=1 since 3 = 1+2; a(5)=2 since 5 = 2+3; a(6)=1 since 6 = 1+2+3; etc.
%p a:= proc(n) local j, k, s; j, k, s:= 1$3;
%p while s<>n do
%p if s<n then k:= k+1; s:= s+k
%p else s:= s-j; j:= j+1 fi
%p od: j
%p end:
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Aug 05 2018
%t a[n_] := Module[{j = 1, k = 1, s = 1}, While[True, If[s == n, Break[]]; If[s < n, k = k+1; s = s+k, s = s-j; j = j+1]]; j];
%t Array[a, 100] (* _Jean-François Alcover_, Mar 12 2019, after _Alois P. Heinz_ *)
%o (PARI) {a(n)=local(A=n);for(j=1,n,for(k=j,n+1,if(n==k*(k-1)/2-j*(j-1)/2,A=j;k=j=2*n+1)));A} /* _Paul D. Hanna_, Oct 28 2011 */
%Y Cf. A000079, A000217, A001227, A002817, A104512, A109814, A212652, A299765.
%K nonn,look
%O 1,2
%A _Reinhard Zumkeller_, Apr 18 2006
|
